CBO与RBO下的IN/EXISTS
发表于:2007-06-07来源:作者:点击数:
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晚上抽空看了看ask tom的RSS,发现两篇应该说很入门的关于IN/EXISTS的文章:http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:953229842074,http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:442029737684 文章很
晚上抽空看了看ask tom的RSS,发现两篇应该说很入门的关于IN/EXISTS的文章:http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:953229842074,http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:442029737684
文章很长,也没有仔细看完,不过有些东西还是很有意思的,动手实验了一下。大家随便看看咯,Tom当初回答问题的环境我已经没办法测试了,在10.1.0.4下做了些测试,CBO和RBO(使用hints)下还是很大区别的。
SQL> select * from scott.emp;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7369 S
MITH CLERK 7902 17-12月-80 800 20
7499 ALLEN SALESMAN 7698 20-2月 -81 1600 300 30
7521 WARD SALESMAN 7698 22-2月 -81 1250 500 30
7566 JONES MANAGER 7839 02-4月 -81 2975 20
7654 MARTIN SALESMAN 7698 28-9月 -81 1250 1400 30
7698 BLAKE MANAGER 7839 01-5月 -81 2850 30
7782 CLARK MANAGER 7839 09-6月 -81 2450 10
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
7839 KING PRESIDENT 17-11月-81 5000 10
7844 TURNER SALESMAN 7698 08-9月 -81 1500 0 30
7876 ADAMS CLERK 7788 23-5月 -87 1100 20
7900 JAMES CLERK 7698 03-12月-81 950 30
7902 FORD ANALYST 7566 03-12月-81 3000 20
7934 MILLER CLERK 7782 23-1月 -82 1300 10
已选择14行。
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=3 Card=14 Bytes=1218)
1 0 TABLE A
CCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=1218)
SQL> create table tmp_emp as select * from scott.emp where ename like 'S%';
表已创建。
SQL> select * from tmp_emp;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-12月-80 800 20
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=3 Card=2 Bytes=174)
1 0 TABLE ACCESS (FULL) OF 'TMP_EMP' (TABLE) (Cost=3 Card=2 Bytes=174)
先测试一下IN:
SQL> select * from tmp_emp where ename in (select ename from scott.emp);
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-12月-80 800 20
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=7 Card=2 Bytes=188)
1 0 HASH JOIN (SEMI) (Cost=7 Card=2 Bytes=188)
2 1 TABLE ACCESS (FULL) OF 'TMP_EMP' (TABLE) (Cost=3 Card=2 Bytes=174)
3 1 TABLE ACCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=98)
表没有做过分析,使用了HASH JOIN (SEMI)。
Tom在回答问题的时候提到:
Select * from T1 where x in ( select y from T2 )is typically processed as:select * from t1, ( select distinct y from t2 ) t2 where t1.x = t2.y;The subquery is evaluated, distincted, indexed (or hashed or sorted) and then joined to the original table -- typically.显然第二个查询是很奇怪的。也正是这个促使我打开了
数据库测试,难道一个IN还需要先DISTINCT一下?没见过IN产生排序操作啊。
SQL> ed
已写入 file afiedt.buf
1 select * from tmp_emp,(select distinct ename from scott.emp) t
2* where tmp_emp.ename=t.ename
SQL> /
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME
---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=8 Card=2 Bytes=188)
1 0 VIEW (Cost=8 Card=2 Bytes=188)
2 1 SORT (UNIQUE) (Cost=8 Card=2 Bytes=202)
3 2 HASH JOIN (Cost=7 Card=2 Bytes=202)
4 3 TABLE ACCESS (FULL) OF 'TMP_EMP' (TABLE) (Cost=3 Card=2 Bytes=188)
5 3 TABLE ACCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=98)
显然不一样了。那么改用RULE模式:
SQL> ed
已写入 file afiedt.buf
1 select /*+ rule */ * from tmp_emp,(select distinct ename from scott.emp) t
2* where tmp_emp.ename=t.ename
SQL> /
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME
---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ----------
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT
7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=HINT: RULE
1 0 MERGE JOIN
2 1 VIEW
3 2 SORT (UNIQUE)
4 3 TABLE ACCESS (FULL) OF 'EMP' (TABLE)
5 1 SORT (JOIN)
6 5 TABLE ACCESS (FULL) OF 'TMP_EMP' (TABLE)
再试试加上索引:
SQL> alter table tmp_emp add constraints pk_tmpemp primary key(ename);
表已更改。
分析一下:
SQL> analyze table tmp_emp compute statistics for table for all indexed columns;
表已分析。
SQL> analyze table scott.emp compute statistics for table for all columns;
表已分析。
SQL> select *
2 from tmp_emp e,(select distinct ename from scott.emp) t
3 where e.ename=t.ename;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME
---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=6 Card=2 Bytes=188)
1 0 VIEW (Cost=6 Card=2 Bytes=188)
2 1 SORT (UNIQUE) (Cost=6 Card=2 Bytes=88)
3 2 NESTED LOOPS (Cost=5 Card=2 Bytes=88)
4 3 TABLE ACCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=70)
5 3 TABLE ACCESS (BY INDEX ROWID) OF 'TMP_EMP' (TABLE) (Cost=1 Card=1 Bytes=39)
6 5 INDEX (UNIQUE SCAN) OF 'PK_TMPEMP' (INDEX (UNIQUE)) (Cost=0 Card=1)
SQL> ed
已写入 file afiedt.buf
1 select /*+rule*/ *
2 from tmp_emp e,(select distinct ename from scott.emp) t
3* where e.ename=t.ename
SQL> /
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME
---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ----------
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT
7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=HINT: RULE
1 0 NESTED LOOPS
2 1 VIEW
3 2 SORT (UNIQUE)
4 3 TABLE ACCESS (FULL) OF 'EMP' (TABLE)
5 1 TABLE ACCESS (BY INDEX ROWID) OF 'TMP_EMP' (TABLE)
6 5 INDEX (UNIQUE SCAN) OF 'PK_TMPEMP' (INDEX (UNIQUE))
SQL> select * from tmp_emp where ename in (select ename from scott.emp);
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
7369 SMITH CLERK 7902 17-12月-80 800 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=5 Card=1 Bytes=44)
1 0 NESTED LOOPS (Cost=5 Card=1 Bytes=44)
2 1 SORT (UNIQUE) (Cost=3 Card=14 Bytes=70)
3 2 TABLE ACCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=70)
4 1 TABLE ACCESS (BY INDEX ROWID) OF 'TMP_EMP' (TABLE) (Cost=1 Card=1 Bytes=39)
5 4 INDEX (UNIQUE SCAN) OF 'PK_TMPEMP' (INDEX (UNIQUE)) (Cost=0 Card=1)
SQL> select /*+rule*/ * from tmp_emp where ename in (select ename from scott.emp);
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
7369 SMITH CLERK 7902 17-12月-80 800 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=HINT: RULE
1 0 NESTED LOOPS
2 1 VIEW OF 'VW_NSO_1' (VIEW)
3 2 SORT (UNIQUE)
4 3 TABLE ACCESS (FULL) OF 'EMP' (TABLE)
5 1 TABLE ACCESS (BY INDEX ROWID) OF 'TMP_EMP' (TABLE)
6 5 INDEX (UNIQUE SCAN) OF 'PK_TMPEMP' (INDEX (UNIQUE))
这次确实接近了很多,总体上看和Tom说的情况差不多。不过RBO这种模式对后边的表对应列选择性低时应该很好,而其他情况恐怕不见得是优化的。下面看看EXISTS:
SQL> select * from tmp_emp t where exists(select null from scott.emp e where t.ename=e.ename);
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
7369 SMITH CLERK 7902 17-12月-80 800 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=5 Card=1 Bytes=44)
1 0 NESTED LOOPS (Cost=5 Card=1 Bytes=44)
2 1 SORT (UNIQUE) (Cost=3 Card=14 Bytes=70)
3 2 TABLE ACCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=70)
4 1 TABLE ACCESS (BY INDEX ROWID) OF 'TMP_EMP' (TABLE) (Cost=1 Card=1 Bytes=39)
5 4 INDEX (UNIQUE SCAN) OF 'PK_TMPEMP' (INDEX (UNIQUE)) (Cost=0 Card=1)
看着和用IN一样哦。
SQL> select /*+rule*/ * from tmp_emp t where exists(select null from scott.emp e where t.ename=e.ename);
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-12月-80 800 20
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=HINT: RULE
1 0 FILTER
2 1 TABLE ACCESS (FULL) OF 'TMP_EMP' (TABLE)
3 1 TABLE ACCESS (FULL) OF 'EMP' (TABLE)
用RBO就不同咯!
SQL> select t.* from tmp_emp t,scott.emp e where t.ename=e.ename;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-12月-80 800 20
7788 SCOTT ANALYST 7566 19-4月 -87 3000 20
执行计划
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=5 Card=2 Bytes=88)
1 0 NESTED LOOPS (Cost=5 Card=2 Bytes=88)
2 1 TABLE ACCESS (FULL) OF 'EMP' (TABLE) (Cost=3 Card=14 Bytes=70)
3 1 TABLE ACCESS (BY INDEX ROWID) OF 'TMP_EMP' (TABLE) (Cost=1 Card=1 Bytes=39)
4 3 INDEX (UNIQUE SCAN) OF 'PK_TMPEMP' (INDEX (UNIQUE)) (Cost=0 Card=1)
这样看来在CBO下,使用内关联、IN、EXISTS很可能得到同一个执行计划(更多的情况就不测试了),优化器会发现三者的语义是相同的;而在较早的RBO下EXISTS采用FILTER而IN相当于对子查询先DISTINCT后关联,内关联则是直接关联就行了。
Tom在他最新的回复中这样说:
Use the RBO and see what you see. way back when I wrote this, that was the "more popular" of the two perhapstoday in 2005, what I said years ago using the RBO does not apply to the CBO. the cbo is smart enough to recognize these two things are effectively the same.IN相当于对子查询先DISTINCT后关联这一条真的没有想通,RBO为什么做这样的事呢?实在是没有普适性,我个人的理解就是设计的时候认为IN后面是跟一个值列表的情况居多,当然先把值算出来,然后NL就可以了,如果后面是一个表里的值那也就一样处理;而且使用IN的时候也许大多是子查询对外层查询的筛选性高,即外层的表较大,而子查询的返回值较少。看看不同情况的不同执行计划,
Oracle在CBO上确实还是花了点心思的,赫赫。
结论:在RBO下,使用IN还是EXISTS需要视情况而定,只要记住
使用IN存在排序和DISTINCT这一步骤应该就不难判断;CBO下优化器会为你选择,怎么写就只是习惯问题了。
原文转自:http://www.ltesting.net